JEE Physics Circular Motion MCQs Set 09

Answer: (b) the acceleration vector is tangent to the circle

Explanation:
1. In uniform circular motion the speed is constant, so there is no tangential acceleration.
2. The acceleration is purely centripetal, pointing to the centre and perpendicular to the velocity, which is tangential. Hence statement (b) is false.

Teacher's Note:
1. Constant speed means zero tangential acceleration.
2. Centripetal acceleration only changes the direction of velocity.

Answer: (c) 24 rad/s

Explanation:
1. \( \omega=\frac{d\theta}{dt}=6t^2 \).
2. At t = 2 s: \( \omega=6\times4=24\ \text{rad/s} \).

Teacher's Note:
1. Angular velocity is the time derivative of angular displacement.
2. The constant 0.5 rad does not matter.

Answer: (a) \( \frac{\pi}{30} \) rad/s

Explanation:
1. The needle turns through 2π rad in 60 s.
2. \( \omega=\frac{2\pi}{60}=\frac{\pi}{30}\ \text{rad/s} \).

Teacher's Note:
1. Use ω = 2π/T with T = 60 s.
2. The minute hand would have ω = π/1800 rad/s.

Answer: (d) \( 2R\sin\frac{\omega t}{2} \)

Explanation:
1. In time t the particle turns through angle ωt, so the displacement is the chord of that arc.
2. \( |\vec d|=2R\sin\frac{\omega t}{2} \).

Teacher's Note:
1. Displacement is the chord, not the arc.
2. For ωt = π the chord equals the diameter 2R.

Answer: (a) \( \frac12 \)

Explanation:
1. Same period means the same angular velocity: \( a=\omega^2r \).
2. \( \frac{a_1}{a_2}=\frac{r_1}{r_2}=\frac24=\frac12 \).

Teacher's Note:
1. With equal ω, acceleration is proportional to radius.
2. With equal speed it would be inversely proportional.

Answer: (a) the centripetal force does not suffer any change in magnitude or direction

Explanation:
1. The centripetal force is \( \frac{mv^2}{r} \), which depends on the square of the speed and always points to the centre.
2. Reversing the direction of motion does not change either, so the force is unchanged.

Teacher's Note:
1. Centripetal force always points inward irrespective of the sense of rotation.
2. The force depends on v², not on the sign of v.

Answer: (c) \( \sqrt{\left(\frac{v^2}{R}\right)^2+a^2} \)

Explanation:
1. Radial (centripetal) component \( \frac{v^2}{R} \) and tangential component a are at right angles.
2. \( a_{net}=\sqrt{\left(\frac{v^2}{R}\right)^2+a^2} \).

Teacher's Note:
1. Tangential and radial accelerations are perpendicular.
2. Add them as vectors, not as numbers.

Answer: (b) momentum changes by 2mv

Explanation:
1. At opposite points the velocities are in opposite directions, so \( \Delta p=mv-(-mv)=2mv \).
2. The speed is the same, so the kinetic energy does not change.

Teacher's Note:
1. Momentum is a vector, kinetic energy a scalar.
2. Opposite velocities give the largest change of momentum.

Answer: (d) \( 8T_0 \)

Explanation:
1. \( T=m\omega^2r \).
2. With ω and r doubled: \( T'=m(2\omega_0)^2(2r)=8T_0 \).

Teacher's Note:
1. Tension supplies the centripetal force.
2. Tension varies as ω² and r.

Answer: (d) 12 m/s

Explanation:
1. \( T_{max}=\frac{mv^2}{r}\Rightarrow v^2=\frac{16\times144}{16}=144 \).
2. So v = 12 m/s.

Teacher's Note:
1. Tension provides the centripetal force.
2. Substitute the maximum tension for the limiting speed.

Answer: (a) \( \frac{v^2b}{Rg} \)

Explanation:
1. Banking angle \( \tan\theta=\frac{v^2}{Rg} \).
2. For a small angle, \( \sin\theta\approx\tan\theta=\frac hb \), so \( h=\frac{v^2b}{Rg} \).

Teacher's Note:
1. The banking angle is small, so sin θ and tan θ are nearly equal.
2. Height difference = width × sin θ.

Answer: (b) \( \theta=\tan^{-1}2 \)

Explanation:
1. 72 km/h = 20 m/s.
2. \( \tan\theta=\frac{v^2}{rg}=\frac{400}{20\times10}=2 \), so \( \theta=\tan^{-1}2 \).

Teacher's Note:
1. The lean angle satisfies tan θ = v²/(rg).
2. Convert km/h to m/s first.

Answer: (d) zero

Explanation:
1. A rigid rod can push as well as pull, so no tension is needed at the top.
2. The particle can just reach the top with zero speed.

Teacher's Note:
1. With a string the minimum speed is √(gl); with a rod it is zero.
2. A rod can supply a force in either direction.

Answer: (b) \( \sqrt{2gl} \)

Explanation:
1. To reach the level of the suspension point, the ball must rise by l with zero speed there (the tension is zero there, so the string does not go slack).
2. \( \frac12mv^2=mgl\Rightarrow v=\sqrt{2gl} \).

Teacher's Note:
1. Use energy conservation up to the horizontal position.
2. At the horizontal position the weight is tangential, so the string stays taut up to that point.

Answer: (c) \( \sqrt{(g\sin\theta)^2+\left(\frac{v^2}{l}\right)^2} \)

Explanation:
1. Tangential acceleration \( a_t=g\sin\theta \) and radial acceleration \( a_r=\frac{v^2}{l} \) are perpendicular.
2. \( a=\sqrt{a_t^2+a_r^2} \).

Teacher's Note:
1. Both tangential and radial components are present when the bob is neither at the extreme nor at the lowest point.
2. At the extreme only g sinθ remains; at the lowest point only v²/l.

Answer: (b) 3 mg

Explanation:
1. Speed at the mean position: \( v^2=2gl \).
2. \( T-mg=\frac{mv^2}{l}=2mg\Rightarrow T=3mg \).

Teacher's Note:
1. Energy conservation gives the speed at the bottom.
2. Tension at the bottom exceeds the weight by the centripetal force.

Answer: (c) \( T_1>T_2 \)

Explanation:
1. \( T=\frac{mv^2}{l}+mg\cos\theta \).
2. At 30° the speed is larger than at 60° (the particle is lower) and cos 30° is larger than cos 60°, so both terms are larger: \( T_1>T_2 \).

Teacher's Note:
1. Tension in a vertical circle decreases as the particle rises.
2. Both the speed and the weight component along the string are larger lower down.

Answer: (b) \( -50\hat i-1.5\hat j \) m/s²

Explanation:
1. Radial component towards the centre (the −x direction): \( \frac{v^2}{r}=\frac{100}{2}=50\ \text{m/s}^2 \).
2. Tangential component along the motion (the −y direction): 1.5 m/s².
3. \( \vec a=-50\hat i-1.5\hat j \).

Teacher's Note:
1. Centripetal acceleration points to the centre.
2. Tangential acceleration is along the velocity when speeding up.

Answer: (a) \( \frac{ML\omega^2}{2} \)

Explanation:
1. The liquid moves in circles, so the net force on it equals \( M\omega^2r_{cm} \), with its centre of mass at \( \frac L2 \).
2. The closed end must supply this force: \( F=\frac{ML\omega^2}{2} \).

Teacher's Note:
1. Use the centre of mass of the liquid for the net centripetal force.
2. The pivot end provides no force on the liquid.

Answer: (c) 24.25 m

Explanation:
1. For a fixed banking angle \( v^2\propto R \).
2. \( R'=R\times(1.1)^2=20\times1.21=24.2\ \text{m}\approx24.25\ \text{m} \).

Teacher's Note:
1. Speed squared is proportional to the radius.
2. A 10% increase in speed is 21% in R.

Answer: (d) 2.5 m/s² and 22.5 m/s²

Explanation:
1. Force varies as \( \omega^2 \), so tripling ω multiplies it by 9: \( 9F-F=60\Rightarrow F=7.5\ \text{N} \).
2. Initial acceleration \( =\frac{7.5}{3}=2.5\ \text{m/s}^2 \), final \( =\frac{67.5}{3}=22.5\ \text{m/s}^2 \).

Teacher's Note:
1. Centripetal force is proportional to ω².
2. Acceleration = force / mass.

Answer: (d) \( h\ge\frac{5r}{2} \)

Explanation:
1. At the top of the loop the minimum speed is \( v^2=gr \).
2. Energy conservation from height h to the top (height 2r): \( mg(h-2r)=\frac12mgr \), so \( h=\frac{5r}{2} \) is the minimum.
3. Any height at or above this works.

Teacher's Note:
1. The condition is an inequality because larger h also completes the loop.
2. Use v² ≥ gr at the top.